Author Archives: Swati

Exercise 1.1

Question 1.
Add the following rational numbers:

Solution:

Question 2.
Add the following rational numbers:

Solution:

Question 3.
Simplify:

Solution:

Question 4.
Add and Express the sum as mixed fraction:

Solution:

Exercise 1.2

Question 1.
Verify commutativity of addition of rational numbers for each of the following pairs of rational numbers :

Solution:

Question 2.
Verify associativity of addition of rational numbers i.e., (A: + y) + z = x + (y + z), when :

Solution:

Question 3.
Write the additive inverse of each of the following rational numbers :

Solution:

Question 4.
Write the negative (additive inverse) of each of the following :

Solution:

Question 5.
Using commutativity and associativity of addition of rational numbers, express ‘iach of the following as a rational number :

Solution:

Question 6.
Re-arrange suitably and find the sum in each of the following :


Solution:

Exercise 1.3

Question 1.
Subtract the first rational number from the second in each of the following :



Solution:

Question 2.
Evaluate each of the following :

Solution:

Question 3.
The sum of two numbers is 59. If one of the numbers is 13, find the other.
Solution:

Question 4.
The sum of two numbers is −13. If one of the numbers is −123, find the other.
Solution:

Question 5.
The sum of two numbers is −43. If one of the number is -5, find the
Solution:

Question 6.
The sum of two rational numbers is -8. If one of the numbers is −157 find the other.
Solution:

Question 7.
What should be added to so as to −78 get 59 ?
Solution:

Question 8.
What number should be added to −511 so as to get 263 ?
Solution:

Question 9.
What number should be added to −57 to get −23 ?
Solution:

Question 10.
What number should be subtracted from −53 to get 56 ?
Solution:

Question 11.
What number should be subtracted from 37 to get 54 ?
Solution:

Question 12.
What should be added to (23+35) to get −1215 ?
Solution:

Question 13.
What should be added to (12+13+15) to get 3 ?
Solution:

Question 14.
What should be subtracted from (34−23) to get −16 ?
Solution:

Question 15.
Fill in the blanks :

Solution:

Exercise 1.4

Question 1.
Subtract the first rational number from the second in each of the following :



Solution:

Question 2.
Evaluate each of the following :

Solution:

Question 3.
The sum of two numbers is 59. If one of the numbers is 13, find the other.
Solution:

Question 4.
The sum of two numbers is −13. If one of the numbers is −123, find the other.
Solution:

Question 5.
The sum of two numbers is −43. If one of the number is -5, find the
Solution:

Question 6.
The sum of two rational numbers is -8. If one of the numbers is −157 find the other.
Solution:

Question 7.
What should be added to so as to −78 get 59 ?
Solution:

Question 8.
What number should be added to −511 so as to get 263 ?
Solution:

Question 9.
What number should be added to −57 to get −23 ?
Solution:

Question 10.
What number should be subtracted from −53 to get 56 ?
Solution:

Question 11.
What number should be subtracted from 37 to get 54 ?
Solution:

Question 12.
What should be added to (23+35) to get −1215 ?
Solution:

Question 13.
What should be added to (12+13+15) to get 3 ?
Solution:

Question 14.
What should be subtracted from (34−23) to get −16 ?
Solution:

Question 15.
Fill in the blanks :

Solution:

Exercise 1.5

Question 1.
Multiply:

Solution:

Question 2.
Multiply:

Solution:

Question 3.
Simplify each of the following and express the result as a rational number in standard form :

Solution:

Question 4.
Simplify :


Solution:

Question 5.
Simplify :

Solution:

Exercise 1.6

Question 1.
Verify the property : x x y = y x x by taking :

Solution:

Question 2.
Verify the property : x x (y x z) = (x x y) x z by taking :

Solution:

Question 3.
Verify the property :xx(y + 2) = xxy + x x z by taking :

Solution:

Question 4.
Use the distributivity of multiplication of rational numbers over their addition to simplify :

Solution:

Question 5.
Find the multiplicative inverse (reciprocal) of each of the following rational numbers :


Solution:

Question 6.
Name the property of multiplication of rational numbers illustrated by the following statements :

Solution:

Exercise 1.7

Question 1.
Divide :

Solution:

Question 2.
Find the value and express as a rational number in standard form :

Solution:

Question 3.
The product of two rational numbers is15. If one of the numbers is -10, find the other.
Solution:

Question 4.
The product of two rational numbers is−98 if one of the numbers is −415, find other.
Solution:

Question 5.
By what number should we multiply −16 so that the product may be −239 ?
Solution:

Question 6.
By what number should we multiply −1528 so that the product may be −57 ?
Solution:

Question 7.
By what number should we multiply −813 so that the product may be 24 ?
Solution:

Question 8.
Bv what number should −34 multiplied in order to produce 23 ?
Solution:

Question 9.
Find (x +y) + (x – y), if

Solution:

Question 10.
The cost of 7 23 metres of rope is Rs 12 34.Find its cost per metre.
Solution:

Question 11.
The cost of 2 13 metres of cloth is Rs. 75 14Find the cost of cloth per metre.
Solution:

Question 12.
By what number should −3316 be divided to get −114 ?
Solution:

Question 13.
Divide the sum of −135 and 127 by the product of −317 and −12.
Solution:

Question 14.
Divide the sum of 6512 and 127 bv their difference.
Solution:

Question 15.
If 24 trousers of equal size can be prepared in 54 metres of cloth, what length of cloth is required for each trouser ?
Solution:

Exercise 1.8

Question 1.
Find a rational number between -3 and 1.
Solution:

Question 2.
Find any five rational number less than 1.
Solution:

Question 3.
Find four rational numbers between −29 and 59 .
Solution:

Question 4.
Find two rational numbers between 15 and 12 .
Solution:

Question 5.
Find ten rational numbers between 14 and 12 .
Solution:

Question 6.
Find ten rational numbers between −25 and 12 .
Solution:

Question 7.
Find ten rational numbers between 35 and 34 .
Solution:

Chapter 12 – Some Applications of Trigonometry Exercise Ex. 12.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

A ladder 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, find the height of the wall.Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21A 1.6 m tall girl stands at a distance of 3.2 m from a lamp-post and casts a shadow of 4.8 m on the ground. Find the height of the lamp-post by using (i) trigonometric ratios (ii) property of similar triangles.Solution 21

Question 22A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30^o to 60^o as he walks towards the building. Find the distance he walked towards the building.Solution 22

Question 23The shadow of a tower standing on a level ground is found to be 40 m longer when Sun’saltitude is 30^o. Find the height of the tower.Solution 23

Question 24From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45^o and 60^o respectively. Find the height of the tower.Solution 24

Let BC be the building, AB be the transmission tower, and D be the point on ground from where elevation angles are to be measured.

Question 25The angles of depression of the top and bottom of 8 m tall building from the top of a multistoried building are 30^o and 45^o respectively. Find the height of the multistoried building and the distance between the two buildings.Solution 25

Question 26A statue, 1.6 m tall, stands on a top of pedestal, from a point on the ground, the angle of elevation of the top of statue is 60^o and from the same point the angle of elevation of the top of the pedestal is 45^o. Find the height of the pedestal.Solution 26

Let AB be the statue, BC be the pedestal and D be the point on ground from where elevation angles are to be measured.

Question 27A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower the angle of elevation of the top of the tower is 60^o. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30^o. Find the height of the tower and the width of the canal.

Solution 27

Question 28From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60^o and the angle of depression of its foot is 45^o. Determine the height of the tower.Solution 28

Question 29As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30^o and 45^o. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.Solution 29

Let AB be the lighthouse and the two ships be at point C and D respectively.

Question 30The angle of elevation of the top of a building from the foot of the tower is 30^o and the angle of elevation of the top of the tower from the foot of the building is 60^o. If the tower is 50 m high, find the height of the building.Solution 30

Question 31From a point on a bridge across a river the angles of depression of the banks on opposite side of the river are 30^o and 45^o respectively. If bridge is at the height of 30 m from the banks, find the width of the river.Solution 31

Question 32Two poles of equal heights are standing opposite each other an either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60^o and 30^o, respectively. Find the height of poles and the distance of the point from the poles.Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

An aeroplane is flying at a height of 210 m. Flying at this height at some instant the angles of depression of two points in a line in opposite directions on both the banks of the river are 45° and 60°. Find the width of the river.

Solution 40

Question 41

The angle of elevation of the top of a chimney from the top of the tower is 60° and the angle of depression of the foot of the chimney from the top of the tower is 30°. If the height of the tower is 40, find the height of the chimney. According to pollution control norms, the minimum height of a smoke emitting chimney should be 100m. State if the height of the above mentioned chimney meets the pollution norms. What value is discussed in this question?Solution 41

Let AC = h be the height of the chimney.

Height of the tower = DE = BC = 40 m

In ∆ABE,

∴AB = BE√3….(i)

In ∆CBE,

tan 30° = 

Substituting BE in (i),

AB = 40√3 × √3

= 120 m

Height of the chimney = AB + BC = 120 + 40 = 160 m

Yes, the height of the chimney meets the pollution control norms.Question 42

Two ships are there in the sea on either side of a light house in such away that the ships and the light house are in the same straight line. The angles of depression of two ships are observed from the top of the light house are 60° and 45°  respectively. If the height of the light house is 200 m, find the distance between the two ships.

Solution 42

Let the ships be at B and C.

In D ABD,

∴ BD = 200 m

In D ADC,

Distance between the two ships = BC = BD + DC

Question 43

The horizontal distance between two poles is 15 m. The angle of depression of the top of the first pole as seen from the top of the second pole is 30°. If the height of the second pole is 24 m, find the height of the first pole.

Solution 43

Here m∠CAB = m∠FEB = 30°.

Let BC = h m, AC = x m

In D ADE,

In D BAC,

Height of the second pole is 15.34 mQuestion 44

The angles of depression of two ships from the top of a light house and on the same side of it are found to be 45^o and 30^o respectively. If the ships are 200 m apart, find the height of the light house.Solution 44

Question 45

The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.Solution 45

Let AQ be the tower and R, S respectively be the points which are 4m, 9m away from base of tower.


As the height can not be negative, the height of the tower is 6 m.Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

From the top of building AB, 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30^o and 60^o respectively. Find

(i) the horizontal distance between AB and CD.

(ii) the height of the lamp post.

(iii) the difference between the heights of the building and the lamp post.Solution 53

Question 54

Solution 54

Question 55

Solution 55

Question 56 (i)

ΔA moving boat is observed from the top of a 150m high cliff moving away from the cliff. The angle of depression of the boat changes from 60˚ to 45˚ in 2 minutes. Find the speed of the boat in m /h.Solution 56 (i)

Let AB be the cliff, so AB=150m.

C and D are positions of the boat.

DC is the distance covered in 2 min.

∠ACB = 60^o and ∠ADB = 45^o

∠ABC = 90^o

In ΔABC,

tan(∠ACB)= 

In ΔABD,

tan(∠ADB)= 

So, DC=BD – BC

  =

Now,

Question 56 (ii)

A man in a boat rowing away from a light house 100 m high takes 2 minutes to change the angle of elevation of the top of the light house from 60^o to 30^o. Find the speed of the boat in metres per minute. (use ) Solution 56 (ii)

Let AB be the lighthouse and C be the position of man initially.

Suppose, a man changes his position from C to D.

As per the question, we obtain the following figure

Let speed of the boat be x metres per minute.

Therefore, CD = 2x

Using trigonometry, we have

Also,

Hence, speed of the boat is 57.8 m.Question 57

From the top of a 120 m high tower, a man observes two cars on the opposite sides of the tower and in straight line with the base of tower with angles of depression as 60˚ and 45˚. Find the distance between the cars. Solution 57

AB is the tower.

DC is the distance between cars.

AB=120m

In ΔABC,

tan(∠ACB) = 

In ΔABD,

tan(∠ADB) = 

So, DC=BD+BC

Question 58

Two points A and B are on the same side of a tower and in the same straight line with its base. The angles of depression of these points from the top of the tower are 60˚ and 45˚ respectively. If the height of the tower is 15 m, then find the distance between these points.Solution 58

Let CD be the tower.

So CD =15m

AB is the distance between the points.

∠CAD = 60^o and ∠CBD = 45^o

∠ADC = 90^o

In ΔADC,

tan(∠CAD)= 

In ΔCBD,

tan(∠CBD)= 

So AB=BD – AD

Question 59

A fire in a building B is reported on telephone to two fire stations P and Q, 20 km apart from each other on a straight road. P observes that the fire is at an angle of 60^o to the road and Q observes that it is at an angle of 45^o to the road. Which station should send its team and how much will this team have to travel?Solution 59

Now, in triangle APB,

sin 60^o = AB/ BP

√3/2 = h/ BP

This gives

h = 14.64 kmQuestion 60

Solution 60

Question 61

A man standing on the deck of a ship, which is 8 m above water level. He observes the angle of elevation of the top of a hill as 60^o and the angle of depression of the base of the hill as 30^o. Calculate the distance of the hill from the ship and the height of the hill.Solution 61

Question 62

Solution 62

Question 63

The angle of elevation of an aeroplane from a point on the ground is 45^o. After a flight of 15 seconds, the elevation changes to 30^o. If the aeroplane is flying at a height of 3000 metres, find the speed of the aeroplane.Solution 63

Question 64

Solution 64

Question 65

Solution 65

Question 66

The angle of elevation of a stationery cloud from a point 2500 m above a lake is 15^o and the angle of depression of its reflection in lake is 45^o. What is the height of the cloud above the lake level? (Use tan 15^o = 0.268)Solution 66

Question 67

Solution 67

Question 68

Solution 68

Question 69

Solution 69

Question 70

Solution 70

Question 71

Solution 71

Question 72

Solution 72

Question 73

Solution 73

Question 74

Solution 74

Question 75

Solution 75

Question 76

From the top of a tower h metre high, the angles of depression of two objects, which are in the line with the foot of the tower are α and β (β > α). Find the distance between the two objects.Solution 76

Question 77

A window of a house is h metre above the ground. From the window, the angles of elevation and depression of the top and bottom of another house situated on the opposite side of the lane are found to be a and b respectively. Prove that the height of the house is h (1 + tan α cot β) metres.Solution 77

Question 78

The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these window are observed to be 60° and 30° respectively. Find the height of the balloon above the ground.Solution 78

Chapter 4 – Triangles Exercise Ex. 4.1

Question 1

Fill in the blanks using correct word given in the brackets:-
(i) All circles are __________. (congruent, similar)
(ii) All squares are __________. (similar, congruent)
(iii) All __________ triangles are similar. (isosceles, equilateral)

(iv) Two triangles are similar, if their corresponding angles are __________. (proportional, equal)

(v) Two triangles are similar, if their corresponding sides are __________. (proportional, equal)
(vi) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)

Solution 1

(i) All circles are similar.
(ii) All squares are similar.
(iii) All equilateral triangles are similar.

(iv) Two triangles are similar, if their corresponding angles are equal.

(v) Two triangles are similar, if their corresponding sides are proportional.
(vi) Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional.Question 2

Write the truth value (T/F) of each of the following statements:

(i) Any two similar figures are congruent.

(ii) Any two congruent figures are similar.

(iii) Two polygons are similar, if their corresponding sides are proportional.

(iv) Two polygons are similar, if their corresponding angles are proportional.

(v) Two triangles are similar if their corresponding sides are proportional.

(vi) Two triangles are similar if their corresponding angles are proportional

Solution 2

(i) False

(ii) True

(iii) False

(iv) False

(v) True

(vi) True

Chapter 4 – Triangles Exercise Ex. 4.2

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 1(ix)

Solution 1(ix)

Question 1(x)

Solution 1(x)

Question 1(xi)

Solution 1(xi)

Question 1(xii)

Solution 1(xii)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 3

Solution 3

Question 4

Solution 4

Question 5

In Fig 7.35, state if PQ || EF.

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Chapter 4 – Triangles Exercise Ex. 4.3

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 2

in Fig. 7.57, AE is the AE is the bisector of the exterior CAD Meeting BC produced in E. If AB = 10 cm, AC = 6 cm and BC = 12 cm, find CE.

Solution 2

Question 3

Solution 3

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 4(v)

Solution 4(v)

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Chapter 4 – Triangles Exercise Ex. 4.4

Question 1(i)

In fig., if AB||CD, find the value of x.

Solution 1(i)

Question 1(ii)

In fig., if AB || CD, find the value of x.

Solution 1(ii)

Question 1(iii)

In fig., AB||CD. If OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4, find x.

Solution 1(iii)

Chapter 4 – Triangles Exercise Ex. 4.5

Question 1

Solution 1

Question 2

In Fig. 7.137, AB || QR. Find the length of PB.

Solution 2

Question 3

In Fig. 7.138, XY || BC. Find the length of XY.

Solution 3

Question 4

In a right angled triangle with sides a and b and hypotenuse c, the altitude drawn on the hypotenuse is x. Prove that ab = cx.Solution 4

We have: 

Question 5

In Fig. 7.140, ABC = 90^o and BD AC. If BD = 8 cm and AD = 4 cm, find CD.

Solution 5

Question 6

In Fig. 7.140, ABC = 90^o and BD AC> If AB = 5.7 cm , BD = 3.8 cm and CD = 5.4 cm, find BC.

Solution 6

Question 7

In fig. 7.141, DE || BC such that AE = (1/4) AC. If AB = 6 cm, find AD

Solution 7

Question 8

Solution 8

Question 9

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using similarity criterion for two triangles, show that  Solution 9

Question 10

If ABC and  AMP are two right triangles, right angled at B and M respectively such that  MAP =  BAC. Prove that

Solution 10

Question 11

A vertical stick 10 cm long casts a shadow 8 cm long. At the same time a tower casts a shadow 30 m long. Determine the height of the tower.Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q. Prove that the rectangle obtained by BP and DQ is equal to the rectangle contained by AB and BC.Solution 17

Question 18

In ABC, AL and CM are the perpendiculars from the vertices A anf C to BC and AB respectively. If AL and CM intersect at O, prove that:

(i) 

(ii) Solution 18

Question 19

ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the mid-points of AB, AC, CD and BD respectively, show that PQRS is a rhombus.Solution 19

Question 20

In an isosceles ABC, the base AB is produced both the ways to P and Q such that AP  BQ = AC². Prove that .Solution 20

Question 21

A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2m/sec. If the lamp is 3.6m above the ground, find the length of her shadow after 4 seconds.Solution 21

Question 22

A vertical stick of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.Solution 22

Question 23

In fig. 7.144, ΔABC is right angled at C and DE AB. prove that ΔABC  ΔADE and hence find the lengths of AE and DE.

Solution 23

Question 24

Solution 24

Question 25

In Fig. 7.144, We have AB||CD||EF, if AB = 6 cm, CD = x cm, EF = 10 cm, BD = 4 cm and DE = y cm, calculate the values of x and y.

Solution 25

Chapter 4 – Triangles Exercise Ex. 4.6

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 2

Solution 2

Question 3

The areas of two similar traingles are 81 cm² and 49 cm² respectively. Find the ratio of their corresponding heights. What is the ratio of their corresponding medians?Solution 3

Question 4

The areas of two similar triangles are 169 cm² and 121 cm² respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.?Solution 4

Question 5

The areas of two similar triangles are 25 cm² and 36 cm² respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other.Solution 5

Question 6

The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.Solution 6

Question 7

Solution 7

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

Question 9

Solution 9

Question 10

Solution 10

Question 11

The areas of two similar triangles are 121 cm² and 64 cm² respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other.Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 19

In Fig.7.180, ΔABC and ΔDBC are two triangles on the same base BC. If AD intersects BC at O,

show that    

Solution 19

Since ABC and DBC are one same base,
Therefore ratio between their areas will be as ratio of their heights.
Let us draw two perpendiculars AP and DM on line BC.


In APO and DMO,
APO = DMO    (Each is90^o)
AOP = DOM          (vertically opposite angles)
OAP = ODM         (remaining angle)
Therefore APO ~  DMO    (By AAA rule)
Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 18

Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3RS. Find the ratio of the areas of triangles POQ and ROS.Solution 18

In trapezium PQRS, PQ || RS and PQ = 3RS.

 … (i)

In ∆POQ and ∆ROS,

∠SOR = ∠QOP … [Vertically opposite angles]

∠SRP = ∠RPQ … [Alternate angles]

∴ ∆POQ ∼ ∆ROS … [By AA similarity criteria]

Using the property of area of areas of similar triangles, we have

Hence, the ratio of the areas of triangles POQ and ROS is 9:1. 

Chapter 4 – Triangles Exercise Ex. 4.7

Question 1

Solution 1

Question 2(i)

The sides of a triangle are a = 7 cm, b = 24 cm and c = 25 cm. Determine whether it is a right triangle.Solution 2(i)

Question 2(ii)

The sides of a triangle are a = 9 cm, b = 16 cm and c = 18 cm. Determine whether it is a right triangle.Solution 2(ii)

Question 2(iii)

The sides of a triangle are a = 1.6 cm, b = 3.8 cm and c = 4 cm. Determine whether it is a right triangle.Solution 2(iii)

Question 2(iv)

The sides of a triangle are a = 8 cm, b = 10 cm and c = 6 cm. Determine whether it is a right triangle.Solution 2(iv)

Question 3

A man goes 15 metres due west and then 8 metres due north. How far is he from the starting point?Solution 3

Question 4

A ladder 17 m long reaches a window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building. Solution 4

Question 5

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.Solution 5

Let CD and AB be the poles of height 11 and 6 m.
Therefore CP = 11 – 6 = 5 m
From the figure we may observe that AP = 12m
In triangle APC, by applying Pythagoras theorem

Therefore distance between their tops = 13 m.Question 6

In an isosceles triangle ABC, AB = AC = 25 cm, BC = 14 cm. Calculate the altitude from A on BC.Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Using pythagoras theorem determine the length of AD terms of b and c shown in Fig. 7.221.

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

In Fig. 7.222, B<90^o and segment AD BC, show that

Solution 17

(i)

Question 18

Solution 18

Question 19

ABD is a right triangle right angled at A and AC  BD. Show that
(i)    AB² = BC . BD
(ii)    AC² = BC . DC
(iii)    AD² = BD . CD

(iv) AB²/ AC² = BD/ DCSolution 19

Question 20

A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?Solution 20

Question 21

Determine whether the triangle having sides (a – 1) cm,  cm and (a + 1) cm is a right angled triangle.Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

In Fig. 7.223, D is the mid-point of side BC and AE  BC. If

Solution 24

Question 25

Solution 25

(i)

Question 26

Solution 26

Question 27

Solution 27

Question 28

An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after Solution 28

Chapter 25 – Probability Exercise Ex. 25.1

Question 1A coin is tossed 1000 times with the following frequencies:

Head : 455, Tail : 545

Compute the probability for each event.Solution 1

Question 2

Two coins are tossed simultaneously 500 times with the following frequencies of different outcomes:

Two heads : 95 times

One tail : 290 times

No head : 115 times

Find the probability of occurrence of each of these events.Solution 2

Question 3

Solution 3

Question 41500 families with 2 children were selected randomly and the following data were recorded:

Number of girls in a family:	0	1	2
Number of families:	211	814	475

If a family is chosen at random, compute the probability that it has:

(i) No girl

(ii) 1 girl

(iii) 2 girls

(iv) at most one girl

(v) more girls than boysSolution 4

Question 5In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays.

Find the probability that on a ball played:

(i) he hits boundary

(ii) he does not hit a boundary.Solution 5

Question 6The percentage of marks obtained by a student in monthly unit tests are given below:

Units Test:	I	II	III	IV	V
Percentage of marks obtained:	69	71	73	68	76

Find the probability that the student gets:

(i) More than 70% marks

(ii) less than 70% marks

(iii) a distinction.Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Following table shows the birth month of 40 students of class IX.
Jan	Feb	March	April	May	June	July	Aug.	Sept.	Oct.	Nov.	Dec.
3	4	2	2	5	1	2	5	3	4	4	4
Find the probability that a student was born in August.

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Chapter 24 – Measures of Central Tendency Exercise Ex. 24.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7Find the mean of first five multiples of 3.Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

The numbers of children in 10 families of a locality are:

2, 4, 3, 4, 2, 0, 3, 5, 1, 1, 5. Find the mean number of children per family.Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

The traffic police recorded the speed (in km/hr) of 10 motorists as 47, 53, 49, 60, 39, 42, 55, 57, 52, 48. Later on an error in recording instrument was found. Find the correct average speed of the motorists if the instrument recorded 5 km/hr less in each case.Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

If M is the mean of x₁, x₂, x₃, x₄, x₅ and x₆ , prove that

(x₁ – M) + (x₂ – M) + (x₃ -M) + (x₄ – M) + (x₅ – M) + (x₆ – M) = 0Solution 19

Question 20

Solution 20

Question 21

Solution 21(i)

(ii)

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Chapter 24 – Measures of Central Tendency Exercise Ex. 24.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Five coins were simultaneously tossed 1000 times and at each toss the number of heads were observed. The number of tosses during which 0, 1, 2, 3, 4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.

No. of heads per toss	No. of tosses
0 1 2 3 4 5	38 144 342 287 164 25
Total	1000

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

x	f	xf
10	17	170
30	f1	30f1
50	32	1600
70	f2	70f2
90	19	1710
	N = 120

Chapter 24 – Measures of Central Tendency Exercise Ex. 24.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9Numbers 50, 42, 35, 2x + 10, 2x – 8, 12, 11, 8 are written in descending order and their median is 25, find x.Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
    29,    32,    48,    50,    x,    x + 2,    72,    78,    84,    95    

Solution 13

Chapter 24 – Measures of Central Tendency Exercise Ex. 24.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4(i)Find the mode of        14,      25,    14,    28,    18,    17,    18,    14,    23,    22,    14,    18Solution 4(i)Arranging the data in an ascending order 14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28  Here observation 14 is having the highest frequency i.e. 4 in given data. So, mode of given data is 14.  

Question 4 (ii)

Solution 4 (ii)

Question 5

The demand of different sizes, as obtained by a survey, is given below:

Size	38	39	40	41	42	43	44	Total
Number of persons (wearing it):	26	39	20	15	13	7	5	125

Find the modal shirt size, as observed from the survey.Solution 5

Chapter 23 – Graphical Representation of Statistical Data Exercise Ex. 23.1

Question 1The following table shows the daily production of T.V. sets in an industry for 7 days a week:

Day	Mon	Tue	Wed	Thurs	Fri	Sat	Sun
No. of T.V. sets	300	400	150	250	100	350	200

Represent the above information by a pictograph.Solution 1The given information can be represented through a pictograph as follows:

Question 2The following table shows the number of Maruti cars sold by five dealers in a particular month:

Dealer	Saya	Bagga Links	D.D. Motors	Bhasin Motors	Competent
Cars sold	60	40	20	15	10

Represent the above information by a pictograph.Solution 2The given information can be represent through a pictograph as follows:

Question 3The population of Delhi State in different census years is as given below:

Census Year	1961	1971	1981	1991	2001
Population In Lakhs	30	55	70	110	150

Represent the above information with the help of bar graph.Solution 3To represent the data by a bar graph, draw horizontal and vertical axes. Mark census year on the horizontal axis and the population on the vertical axis.

Question 4Read the bar graph shown in figure and answer the following questions:

(i) What is the information given by the bar graph?

(ii) How many tickets of Assam state lottery were sold by the agent?

(iii) Of which state, were the maximum number of tickets sold?

(iv) State whether true or false.

      The maximum number of tickets sold is three the minimum number of tickets sold.

(v) Of which state were the minimum number of tickets sold?

Solution 4(i) The given bar graph represents the number of tickets of different state lotteries sold by an agent on a day.

(ii) The number of tickets of Assam state lottery sold by the agent is 40.

(iii) Haryana

(iv) The minimum number of tickets sold = 20

      The maximum number of tickets sold = 100

So, the given statement is false.

(v) RajasthanQuestion 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Read the following bar graph and answer the following questions:

(i) What information is given by the bar graph?

(ii) Which state is the largest producer of rice?

(iii) Which state is the largest producer of wheat?

(iv) Which state has total production of rice and wheat as its maximum?

(v) Which state has total production of wheat and rice minimum?

Solution 9

(i) It gives information regarding rice and wheat production in various states of India.

(ii) W.B is the largest produer of rice.

(iii) U.P is the largest producer of wheat.

(iv) The total production of rice and wheat is maximum in U.P.

(v) The total production of rice and wheat is minimum in Maharashtra.Question 10

Solution 10

Question 11

Solution 11

Question 13Read the bar graph given in fig., and answer the following questions:

(i) What information is given by the bar graph?

(ii) What was the crop-production of rice in 1970-71?

(iii) What is the difference between the maximum and minimum production of rice?

Solution 13(i) It gives information regarding the production of rice crop in India in different years.

(ii) The crop-production of rice in 1970-71 = 42.5 lakh tonnes.

(iii) The difference between the maximum and minimum production of rice = 55 – 22 = 33 lakh tonnes.Question 14Read the bar graph given fig., and answer the following questions:

(i) What information does it give?

(ii) In which part the expenditure on education is maximum in 1980?

(iii) In Which part the expenditure has gone up from 1980 to 1990?

(iv) In which part the gap between 1980 and 1990 is maximum?

Solution 14(i) It gives the information about the public expenditure on education by various state subcontinents.

(ii) In Africa the expenditure on edcation is maximum in 1980.

(iii) In East Africa the expenditure has gone by from 1980 to 1990.

(iv) In Africa the gap between 1980 and 1990 is maximum.Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Chapter – Exercise

SolutionSolution

Chapter 23 – Graphical Representation of Statistical Data Exercise Ex. 23.2

Question 1Explain the reading and interpretation of bar graphs.Solution 1The first step in reading a bar graph is to know what it represents or what is the information given by it. For this, we read the captions which are generally written just below the horizontal line (x-axis) and adjacent to vertical line (y-axis). After knowing that what does a bar graph represent, we read the scale so that we can know the precise values in the given data.

After reading a bar graph one must be able to draw certain condusions from it. Drawing some condusions from a given bar graph means interpretation of the bar graph.Question 2

Solution 2

Question 3The following bar graph shows the results of an annual examination in a secondary school.

Read the bar graph (fig.,) and choose the correct alternative in each of the following

(i) The pair of classes in which the results of boys and girls are inversely proportional are:

(a) VI, VIII

(b) VI, IX

(c) VIII, IX

(d) VIII, X

(ii) The class having the lowest failure rate of girls is

(a) VII

(b) X

(c) IX

(d) VIII

(iii) The class having the lowest pass rate of students is

(a) VI

(b) VII

(c) VIII

(d) IXSolution 3(i) (b) VI, IX

(ii) (a) VII

(iii) (b) VIIQuestion 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7The following data gives the amount of loans (in croroes of rupees) disbursed by a bank during some years:

Year	1992	1993	1994	1995	1996
Loan (in crores of rupees)	28	33	55	55	80

(i) Represent the above data with the help of a bar graph.

(ii) With the help of the bar graph, indicate the year in which amount of loan is not increased over that of the preceding year.Solution 7

(i) Bar graph for the given data is as follows:

(ii) In 1995, the amount of loan is not increased over that of the preceeding year.Question 8The following table shows the interest paid by a company (in lakhs):

Year	1995-96	1996-97	1997-98	1998-99	1999-2000
Interest (in lakhs of rupees)	20	25	15	18	30

Draw the bar graph to represent the above information.Solution 8

Question 9The following data shows the average age of men in various countries in a certain year:

Country	India	Nepal	China	Pakistan	U.K.	U.S.A.
Average age (in years)	55	52	60	50	70	75

Represent the above information by a bar graph.Solution 9

Question 10The following data gives the production of foodgrains (in thousand tonnes) for some years:

Year	1995	1996	1997	1998	1999	2000
Production (in thousand tonnes)	120	150	140	180	170	190

Represent the above data with the help of a bar graph.Solution 10

Question 11

Solution 11(i)

(ii) It is seen that the height of bar corresponding to year 1994 is the highest. Hence, the amount of manure manufactured by the company was maximum in 1994.

(iii) It is seen that the manure production decreased in the year 1995 and 1997.Question 12

Solution 12The bar graph of the given bata:


The course where estimated requirement is least is DCEQuestion 13

Solution 13

Question 14The investment (in ten crores of rupees) of Life Insurance Corporation of India in different sectors are given below:

Sectors	Investment (in ten crores of rupees)
Central Government Securities State Government Securities Securities guaranteed by the Government Private Sectros Socially oriented sectors (Plan) Socially oriented sectors (Non-Plan)	45 11 23 18 46 11

Represent the above data with the help of a bar graph.Solution 14

Question 15The following data gives the value (in crores of rupees) of the Indian export of cotton textiles for different years:

Years	1982-83	1983-84	1984-85	1985-86	1986-87
Value of Exports of Cotton Textiles (in crores of rupees)	300	325	475	450	550

Represent the above data with the help of a bar graph. Indicate with the help of a bar graph the year in which the rate of increase in exports is maximum over the preceding years.Solution 15The bar graph of the given data:

In 1986-87 the rate of increases in exports is maximum over the precending year.Question 16

Solution 16

It is seen from the graph that the quantity of goods carried in the years 1950-51 and 1965-66 are 20 crores tones and 9 crores tones. Clearly 20 is more than twice of 9. Hence, the statement is true.Question 17

Solution 17

Question 18

Solution 18

Chapter 23 – Graphical Representation of Statistical Data Exercise Ex. 23.3

Question 1Construct a histogram for the following data:

Monthly School fee (in Rs.)	30-60	60-90	90-120	120-150	150-180	180-210	210-240
No. of Schools	5	12	14	18	10	9	4

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 7

Solution 7

Question 8The monthly profits (in Rs.) of 100 shops are distributed as follows:

Profits per shop:	0-50	50-100	100-150	150-200	200-250	250-300
No. of shops:	12	18	27	20	17	6

Draw a histogram for the data and show the frequency polygon for it.Solution 8

Chapter 22 – Tabular Representation of Statistical Data Exercise Ex. 22.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4Why do we group data?Solution 4The data obtained in original form are called raw data. Raw data does not give any useful information and is rather confusing to mind. Data is grouped so that it becomes understandable and can be interpreted. We form groups according to various characteristics. After grouping the data, we are in a position to make calculations of certain values which will help us in describing and analysing the data.Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9The final marks in mathematics of 30 students are as follows:

53,61,48,60,78,68,55,100,67,90,75,88,77,37,84,58,60,48,62,56,44,58,52,64,98,59,70,39,50,60

(i)

Group	I(30-39)	II(40-49)	III(50-59)	IV(60-69)	V(70-79)	VI(80-89)	VII(90-99)	VIII(100-109)
Observations	37, 39	44, 48, 48	50, 52, 53, 55, 56, 58, 58, 59	60, 60, 60, 61, 62, 64, 67, 68	70, 75, 77, 78	84, 88	90, 98	100

(ii) Highest score = 100

(iii) Lowest score = 37

(iv) Range = 100 – 37 = 63

(v) If 40 is the pass mark, 2 students have failed.

(vi) 8 students have scored 75 or more.

(vii) Observations 51, 54, 57 between 50 and 60 have not actually appeared.

(viii) 5 students have scored less than 50.

Question 10

Solution 10

Question 11The number of runs scored by a cricket player in 25 innings are as follows:

26,35,94,48,82,105,53,0,39,42,71,0,64,15,34,67,0,42,124,84,54,48,139,64,47.

(i) Rearrange these runs in ascending order.

(ii) Determine the player’s highest score.

(iii) How many times did the player not score a run?

(iv) How many centuries did he score?

(v) How many times did he score more than 50 runs?Solution 11The numbers of runs scored by a player in 25 innings:

26, 35, 94, 48, 82, 105, 53, 0, 39, 42, 71, 0, 64, 15, 34, 67, 0, 42, 124, 84, 54, 48, 139, 64, 47.

(i) Runs in ascending order:- 0,0,0,15,26,34,35,39,42,42,47,48,48,53,54,64,64,67,71,82,84,94,105,124,139

(ii) The highest score = 139

(iii) The player did not score any run 3 times.

(iv) He scored 3 centuries.

(v) He scored more than 50 runs 12 times.Question 12

Solution 12

Question 13Write the class size and class limits in each of the following

(i) 104, 114, 124, 134, 144, 154, and 164

(ii) 47, 52, 57, 62, 67, 72, 82, 87, 92, 97 and 102

(iii) 12.5, 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5Solution 13(i)

(ii)

(iii)

Question 14

Solution 14

Number of children	Tally marks	Number of families
0		5
1	ll	7
2	ll	12
3		5
4	l	6
5	lll	3
6	lll	3

Question 15

Solution 15

Marks	Tally marks	Frequency
20 – 30	l	1
30 – 40	lll	3
40 – 50		5
50 – 60	lll	8
60 – 70	lll	8
70 – 80	llll	9
80 – 90	llll	4
90 – 100	ll	2
		Total = 40

Question 16

The heights (in cm) of 30 students of class IX are given below:

155, 158,154, 158, 160, 148, 149, 150, 153, 159, 161, 148, 157, 153, 157, 162, 159, 151, 154, 156, 152, 156, 160, 152, 147, 155, 163, 155, 157, 153.

Prepare a frequency distribution table with 160-164 as one of the class intervals.Solution 16

Heights (in cm)	Tally marks	Frequency
145 – 149	llll	4
150 – 154	llll	9
155 – 159	ll	12
160 – 164		5
		Total = 30

Question 17

Solution 17

Height (in cm)	Tally marks	Frequency
800 – 810	lll	3
810 – 820	ll	2
820 – 830	l	1
830 – 840	lll	8
840 – 850		5
850 – 860	l	1
860 – 870	lll	3
870 – 880	l	1
880 – 890	l	1
890 – 900		5
		Total = 30

Question 18

Solution 18

Maximum temperature (in degree Celsius)	Tally marks	Frequency
20.0 – 21.0	l	6
21.0 – 22.0		5
22.0 – 23.0	llll	9
23.0 – 24.0		5
24.0 – 25.0	lll	3
25.0 – 26.0	ll	2
		Total = 30

Question 19

Solution 19

Monthly wages (in rupees)	Tally marks	Frequency
210 – 230	llll	4
230 – 250	llll	4
250 – 270		5
270 – 290	lll	3
290 – 310	ll	7
310 – 330		5
		Total = 28

Question 20

The blood groups of 30 student of Class VIII are recoded as follows:

A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,

A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.

Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?Solution 20Here 9 students  have blood groups  A, 6 as B, 3 as AB and 12 as O.
So, the table representing the data is as follows:  

Blood group	Number of students
A	9
B	6
AB	3
O	12
Total	30

As 12 students have the blood group O and 3 have their blood group as AB. Clearly, the most common blood group among these students is O and the rarest blood group among these students is AB.Question 21Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:
              0    1    2    2    1    2    3    1    3    0
              1    3    1    1    2    2    0    1    2    1
              3    0    0    1    1    2    3    2    2    0

Prepare a frequency distribution table for the data given above.   

Solution 21By observing the data given above following frequency distribution table can be constructed

Number of heads	Number of times (frequency)
0	6
1	10
2	9
3	5
Total	30

Question 22Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:

     1    6    2      3    5    12      5    8      4     8
    10   3    4      12   2     8      15   1    17     6
     3    2    8      5    9      6      8    7    14    12

    (i)    Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5 – 10.
    (ii)    How many children watched television for 15 or more hours a week? 

Solution 22(i) Class intervals will be 0 – 5, 5 – 10, 10 -15…..
    The grouped frequency distribution table is as follows:

Hours	Number of children
0 – 5	10
5 – 10	13
10 – 15	5
15 – 20	2
Total	30

(ii) The number of children, who watched TV for 15 or more hours a week
        is 2 (i.e. number of children in class interval 15 – 20).

Question 23

Solution 23

Since first class interval is -19.9 to -15

Frequency distribution with lower limit included and upper limit excluded is:

Temperature	Tally marks	Frequency
-19.9 to -15	ll	2
-15 to -10.1	ll	7
-10.1 to -5.2		5
-5.2 to -0.3	llll	4
-0.3 to 4.6	ll	17
Total		35

Chapter 22 – Tabular Representation of Statistical Data Exercise Ex. 22.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4Following are the the ages of 360 patients getting medical treatment in a hospital on a day:

Age (in years):	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70
No. of Patients:	90	50	60	80	50	30

Construct a cumulative frequency distribution.Solution 4

Age (in years):	No. of patients	Age (in years)	Cumulative frequency
10 – 20	90	Less than 20	90
20 – 30	50	Less than 30	140
30 – 40	60	Less than 40	200
40 – 50	80	Less than 50	280
50 – 60	50	Less than 60	330
60 – 70	30	Less than 70	360
	N = 360

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8The following cumulative frequency distribution table shows the daily electricity consumption (in kW) of 40 factories in an industrial state:

Consumption (in KW)	No. of Factories
Below 240 Below 270 Below 300 Below 330 Below 360 Below 390 Below 420	1 4 8 24 33 38 40

(i) Represent this as a frequency distribution table.

(ii) Prepare a cumulative frequency table.Solution 8(i)

Consumption (in kW)	No. of Factories	Class interval	Frequency
Below 240	1	0 – 240	1
Below 270	4	240 – 270	4 – 1 = 3
Below 300	8	270 – 300	8 – 4 = 4
Below 330	24	300 – 330	24 – 8 = 16
Below 360	33	330 – 360	33 – 24 = 9
Below 390	38	360 – 390	38 – 33 = 5
Below 420	40	390 – 420	40 – 38 = 2

(ii)

Class interval	Frequency	Consumption (in kW)	No. of factories
0 – 240	1	More than 0	40
240 – 270	3	More than 270	40 – 1 = 39
270 – 300	4	More than 270	39 – 3 = 36
300 – 330	16	More than 300	36 – 4 = 32
330 – 360	9	More than 330	32 – 16 = 16
360 – 390	5	More than 360	16 – 9 = 7
390 – 420	2	More than 390	7 – 5 = 2
		More than 420	2 – 2 = 0
	N = 40

Question 9

Solution 9

Chapter 21 – Surface Areas and Volume of a Sphere Exercise Ex. 21.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 4 per 100 cm².Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.Solution 9

Question 10

A hemi-spherical dome of a building needs to be painted. If the circumference of the base of the dome is 17.6 cm, find the cost of painting, if given the cost of painting is Rs 5 per 100 cm².Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in the given figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm² and black paint costs 5 paise per cm². 

                                      Solution 13

Chapter 21 – Surface Areas and Volume of A Sphere Exercise Ex. 21.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9If the radius of a sphere is doubled, what is the ratio of the volume of the first sphere to that of the second sphere?Solution 9

Question 10

A vessel in the form of a hemispherical bowl is full of water. Its contents are emptied in a right circular cylinder. The internal radii of the bowl and the cylinder are 3.5 cm and 7 cm respectively. Find the height to which the water will rise in the cylinder.Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

The diameter of a coper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 108 m, find its diameter.Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

A cube of side 4 cm contained a sphere touching its sides. Find the volume of the gap in between.Solution 21

Question 22

A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.Solution 22Inner radius (r₁) of hemispherical tank  = 1 m
     Thickness of hemispherical tank       = 1 cm = 0.01 m
Outer radius (r₂) of hemispherical tank = (1 + 0.01) m = 1.01 mVolume of iron used to make the tank  = 

                                                            Question 23

A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule?Solution 23Radius (r) of capsule
Volume of spherical capsule

Thus, approximately 22.46 mm³ of medicine is required to fill the capsule.

Question 24

The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?Solution 24    Let diameter of earth be d. So, radius earth will be  .
    Then, diameter of moon will be  . So, radius of moon will be  .
    Volume of moon =    
    Volume of earth =   

    Thus, the volume of moon is  of volume of earth.

Question 25

Solution 25

Question 26

A cylinderical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9 cm. What is the radius of the ball?Solution 26

Question 27

A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. (Use  = 22/7)Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Chapter 20 – Surface Areas and Volume of A Right Circular Cone Exercise Ex. 20.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6Find its curved surface area of a cone with base radius 5.25 cm and slant height 10 cm.Solution 6

Question 7Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.Solution 7

Question 8

Solution 8

Question 9The curved surface area of a cone is 4070 cm² and its diameter is 70 cm. What is its slant height? (Use  = 22/7).Solution 9

Question 10

Solution 10

Question 11A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.Solution 11

Question 12Find the ratio of the curved surface area of two cones if their diameters of the bases are equal and slant heights are in the ratio 4:3.Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find the radius of the base and total surface area of the cone.Solution 15(i)    Slant height of cone = 14 cm
    Let radius of circular end of cone be r.
    CSA of cone = 

    Thus, the radius of circular end of the cone is 7 cm.

(ii)    Total surface area of cone = CSA of cone + Area of base
                     =

Thus, the total surface area of the cone is 462 .

Question 16The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs 210 per 100 m².Solution 16

Question 17A conical tent is 10 m high and the radius of its base is 24 m. Find slant height of the tent. If the cost of 1 m² canvas is Rs 70, find the cost of the canvas required to make the tent,Solution 17

Question 18

Solution 18

Question 19

What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use  = 3.14).

Solution 19Height (h) of conical tent = 8 m
    Radius (r) of base of tent = 6 m
    Slant height (l) of tent = 
    CSA of conical tent =  = (3.14  6  10)  = 188.4 

Let length of tarpaulin sheet required be L.
As 20 cm will be wasted so, effective length will be (L – 0.2 m)
Breadth of tarpaulin = 3 m
Area of sheet = CSA of tent
    [(L – 0.2 m)  3] m = 188.4 
    L – 0.2 m = 62.8 m
    L = 63 m

    Thus, the length of the tarpaulin sheet will be 63 m.Question 20

Solution 20

Question 21

A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, show that the radius of each to the height of each is 3:4.Solution 21

Question 22

Solution 22

Question 23

Solution 23

Chapter 20 – Surface Areas and Volume of A Right Circular Cone Exercise Ex. 20.2

Question 1Find the volume of the right circular cone with

(i) radius 6 cm, height 7 cm

(ii) radius 3.5 cm, height 12 cm

(iii) height 21 cm and slant height 28 cm.Solution 1(i)    Radius (r) of cone = 6 cm
       Height (h) of cone = 7 cm
       Volume of cone 

(ii)   Radius (r) of cone = 3.5 cm
       Height (h) of cone = 12 cm
       Volume of cone  
(iii)

Question 2Find the capacity in litres of a conical vessel with
(i)    radius 7 cm, slant height 25 cm
(ii)   height 12 cm, slant height 13 cm

Solution 2(i)    Radius (r) of cone = 7 cm
       Slant height (l) of cone = 25 cm
       Height (h) of cone  
       Volume of cone  
       Capacity of the conical vessel =  litres= 1.232 litres(ii)    Height (h) of cone = 12 cm
        Slant height (l) of cone = 13 cm        
        Radius (r) of cone 
        Volume of cone   = 314.28 cm³
        Capacity of the conical vessel = litres =  litres.

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9A heap of wheat is in the form of a cone of diameter 9 m and height is 3.5 m. Find its volume. How much canvas cloth is required to just cover the heap? (use  = 3.14).Solution 9

Question 10

Solution 10

Question 11A right angled triangle of which the sides containing the right angle are 6.3 cm and 10 cm in length, is made to turn round on the longer side. Find the volume of the solid, thus generated. Also, find its curved surface area.
Solution 11

Question 12

Solution 12

Question 13The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find

(i) height of the cone

(ii) slant height of the cone

(iii) curved surface area of the coneSolution 13(i)    Radius of cone =   =14 cm
       Let height of cone be h.
       Volume of cone = 9856 cm³
        h = 48 cm       Thus, the height of the cone is 48 cm. (ii)   Slant height (l) of cone  
       Thus, the slant height of the cone is 50 cm. (iii)    CSA of cone = rl = Question 14

A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?Solution 14Radius (r) of pit =
Depth (h) of pit = 12 m
Volume of pit =Capacity of the pit = (38.5  1) kilolitres = 38.5 kilolitres

Question 15Monica has a piece of canvas whose area is 551 m². She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and the wastage incurred while cutting, amounts to approximately 1 m², find the volume of the tent that can be made with it.Solution 15

Chapter 19 – Surface Areas and Volume of a Circular Cylinder Exercise Ex. 19.1

Question 1Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.

Solution 1

Question 2In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.Solution 2 Height (h) of cylindrical pipe = Length of cylindrical pipe = 28 m
  Radius (r) of circular end of pipe =  cm = 2.5 cm = 0.025 m
  CSA of cylindrical pipe =  = 4.4 
    Thus, the area of radiating surface of the system is 4.4 m² or 44000 cm².

Question 3A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs.12.50 per m².Solution 3Height of the pillar = 3.5 m
Radius of the circular end of the pillar = cm = 25 cm  = 0.25 m
CSA of pillar =  = 
Cost of painting 1  area = Rs 12.50
Cost of painting 5.5  area = Rs (5.5 12.50) = Rs 68.75

Thus, the cost of painting the CSA of pillar is Rs 68.75.

Question 4It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same?   Solution 4Height (h) of cylindrical tank = 1 m.
    Base radius (r) of cylindrical tank =  = 70 cm = 0.7 m
Area of sheet required = total surface area of tank = 

So, it will require 7.48 of metal sheet.

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find (i)    Its inner curved surface area,
 (ii)    The cost of plastering this curved surface at the rate of Rs 40 per m²

Solution 8Inner radius (r) of circular well = 1.75 m
Depth (h) of circular well = 10 m (i) Inner curved surface area = 

          = (44 x 0.25 x 10) 
          = 110 m²(ii) Cost of plastering 1 m² area = Rs 40                    
    Cost of plastering 110 m² area = Rs (110 x 40) = Rs 4400

Question 9The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?   Solution 9Radius of circular end of cylindrical penholder = 3 cm
Height of penholder = 10.5 cm
Surface area of 1 penholder = CSA of penholder + Area of base of     SA of 1 penholder =  + 
Area of cardboard sheet used by 1 competitor = 
    Area of cardboard sheet used by 35 competitors
 = 7920 cm2
    Thus, 7920 cm² of cardboard sheet will be required for the competition.

Question 10

Solution 10

Question 11

Twenty cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50 m and height is 4 m. What will be the cost of cleaning them at the rate of Rs 2.50 per square metre?Solution 11

Question 12

Solution 12

Question 13

The total surface area of a hollow metal cylinder open at both ends of external radius 8 cm and height 10 cm is 338 cm². Taking r to be inner radius, obtain an equation in r and use it to obtain the thickness of the metal in the cylinder.Solution 13

Question 14

Find the lateral or curved surface area of a cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high. How much steel was actually used, if  of the steel actually used was wasted in making the closed tank?Solution 14Height (h) cylindrical tank = 4.5 m
Radius (r) of circular end of cylindrical tank =m = 2.1m
    (i)    Lateral or curved surface area of tank = 
                                    = 
                                    = 59.4 m2            

    (ii)    Total surface area of tank = 2 (r + h)
              = 
              = 87.12 m²

    Let A m² steel sheet be actually used in making the tank.


Thus, 95.04  steel was used in actual while making the tank.    

Chapter 19 – Surface Areas and Volume of a Circular Cylinder Exercise Ex. 19.2

Question 1A soft drink is available in two packs –

(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and(ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?   Solution 1The tin can will be cuboidal in shape.Length (l) of tin can = 5 cm
Breadth (b) of tin can = 4 cm
Height (h) of tin can = 15 cm
Capacity of tin can = l  b  h = (5  4  15) cm³ = 300 cm³Radius (R) of circular end of plastic cylinder = Height (H) of plastic cylinder = 10 cmCapacity of plastic cylinder = R²H  ==385 cm³Thus, the plastic cylinder has greater capacity.
Difference in capacity = (385 – 300) cm³ = 85 cm³

Question 2The pillars of a temple are cylindrically shaped. If each pillar has a circular base of radius 20 cm and height 10m. How much concrete mixture would be required to build 14 such pillars?

Solution 2

Question 3The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. find the mass of the pipe, if 1 cm³ of wood has a mass of 0.6 g.Solution 3

Question 4If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, then find
(i) radius of its base    (ii) its volume. (Use  = 3.14)Solution 4(i)    Height (h) of cylinder = 5 cm
        Let radius of cylinder be r.
        CSA of cylinder = 94.2 cm²
        2rh = 94.2 cm²
        (2  3.14  r  5) cm = 94.2 cm²
        r = 3 cm (ii)    Volume of cylinder = r²h = (3.14  (3)² 5) cm³ = 141.3 cm³

Question 5The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?Solution 5Let radius of the circular ends of the cylinder be r.
Height (h) of the cylindrical vessel = 1 m
Volume of cylindrical vessel = 15.4 litres = 0.0154 m³

Total  Surface area of vessel = 2 r(r+h)

                                            Thus, 0.4708 m² of metal sheet would be needed to make the cylindrical vessel.   Question 6A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?Solution 6Radius (r) of cylindrical bowl = cm = 3.5 cm
Height (h) up to which the bowl is filled with soup = 4 cmVolume of soup in 1 bowl = r²h= 
Volume of soup in 250 bowls = (250  154) cm³ = 38500 cm³ = 38.5 litres

Thus, the hospital will have to prepare 38.5 litres of soup daily to serve 250 patients.

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

A cylinderical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 32 cm x 22 cm x 14 cm. Find the rise in the level of the water when the solid is completely submerged.Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to around it to a width of 21 m to form an embankment. Find the height of the embankment.Solution 27

Question 28

Solution 28

Question 29

Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 meters per second into cylindrical tank. The water is collected in a cylindrical vessel radius of whose base is 60 cm. Find the rise in the level of water in 30 minutes?Solution 29

Question 30

Solution 30

Question 31

The sum of the radius of the base and height of a solid cylinder is 37 m. If the total surface area of the solid cylinder is 1628 m². Find the volume of the cylinder.Solution 31

Question 32A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.Solution 32